Problem: What is the inverse of the function $g(x)=\dfrac{9x+4}{x-7}$ ? $g^{-1}(x) =$
Let's start by replacing $g(x)$ with $y$. $y=\dfrac{9x+4}{x-7}$ Now let's swap $x$ and $y$ and solve for $y$. $\dfrac{9y+4}{y-7}=x$ [Why do we swap x and y?] $\begin{aligned} \dfrac{9y+4}{y-7}&=x \\\\ 9y+4&=x(y-7) \\\\ 9y+4&=xy-7x \\\\ 9y-xy&=-7x-4 \\\\ y(9-x)&=-7x-4 \\\\ y&=\dfrac{-7x-4}{9-x} \end{aligned}$ In conclusion, this is the inverse function: $g^{-1}(x)=\dfrac{-7x-4}{9-x}$ [I saw someone solve this problem by originally solving for x. Were they wrong?]